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**Posts by Hixy**

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**Hixy**- Replies: 3

The problem is as follows:

Given two matrices, **A** and **B** (see below),

a) State for any *a* all eigenvalues for both **A** and **B**.

b) Give for both A and B the algebraic and geometric multiplicities for all of the eigenvalues.

c) For which values of *a* are **A** and **B**, respectively, similar to a diagonal matrix?

d) For which values of *a* are **A** and **B** mutually similar?

Note: By "similar" I mean as in similarity between matrices (http://en.wikipedia.org/wiki/Similar_matrix)

My answers:

a) By finding the characteristic polynomial, i.e.

Is this correct? Am I missing something about the *a* 'complexity'?

b) Here it's simply just finding the eigenvectors and giving their number for the geometric multiplicity for

and .c) Matrix **B** can only be similar to the 3x3 diagonal matrix with the entries 1, -1 and 1. Thus *a* in **A** must be -1. Right?

d) What is the best way to check for matrices being similar? I know that similar matrices can be thought of as describing the same linear map with respect to different bases. Thus I've checked for rank, determinant, trace, eigenvalues, the characteristic polynomial and given the respective values for *a* for each of these to equal each other for **A** and **B**. Is this the way to go?

**Hixy**- Replies: 1

Last time you guys were able to see through the (to me) somewhat cluttered use of integrals and differentiation (I'm still very grateful for your amazing help). This time my question is similar. I'm currently trying to understand the following derivation of *Lagrange's equations* from Landau and Lifshi-tz' amazing first volume on mechanics:

The function for the action

is given. are the coordinates and the velocities, respectively, for a system of particles.

Then the change in

When this difference is expanded in powers of and in the integrand, the leading terms are of the first order

or, affecting the variation (2),

The conditions show that the integrated term is zero. There remains an integral which must vanish for all values of . This can be so only if the integrand is zero identically. Thus we have

What is meant by the sentence at (1)? Which calculations aren't shown here?

How is this translated to affect the variation at (2)?

How did we get to (3)?

Edit: Having studied this further the last couple of days, I've realized what a stupid question it was. I'm gonna read up on calculus of variations and understand this fully!

Just realized I might have something. Hang on.

Edit:

Okay I got it.

Rewrite the original expression as

is an arithmetic series (=the sum of an arithmetic progression). The nth term can be expressed as

, with

By double counting, the entire sum can be expressed as

In (2) the sum is expressed in terms of , and in (3) in terms of . Counting the same set twice is indeed a useful technique!

Adding both sides and dividing by 2 gives

Since , it follows

Finding the functions for and inserting in (1) reduces the entire thing to

Arithmetic series wiki: http://en.wikipedia.org/wiki/Arithmetic_progression

Double counting wiki: http://en.wikipedia.org/wiki/Double_counting_(proof_technique)

**Hixy**- Replies: 1

First the warm-up problem:

My reasoning:

Is this the correct way to look at it? Can the function be found somehow?

Now to the problem:

Any hints as to how I can progress with this? I've tried isolating f(n) and find the starting point of the function, but that is as far as my thought process has brought me at the moment.

**Hixy**- Replies: 1

I need to imprint the following terms on my mind:

1) Set

2) Abelian groups

3) Groups

4) Fields

5) Vector spaces

6) Subspaces

If you had to explain them concretely, what would you say?

is what I've understood. Before seeing my tries at defining the terms, could you write yours? It could happen, unknowningly, that your definitions got influenced by mine in one way or the other.

I hope you can help. Getting a grasp of this should be quite important to understand linear algebra successfully.

Hahaha, yeah.. That is indeed abusive! Thanks for the quick and illustrative reply

**Hixy**- Replies: 3

In a physics problem, we end up with the expression

The wording of the solution is like so:

"Expanding this to first order in

What does this mean and how was this 'expansion' performed?

There are still a few english mathematical terms that I'm unfamiliar with. I apologize if this question is too trivial.

Hi

Just wanted to add another way of solving the problem with matrices which can make the problem quite simple. You can find it

.Hopefully that gave you a slightly different perspective on the problem.

Steps:

1. Dissect the problem and extract the important information from each sentence (this is actually two steps). This could be done as follows (some parts are underlined to add emphasis):

"A bus travels 200 km at constant speed":

"A car traveling 10 km/h faster than the bus completes the same distance in 1 hour less than the bus."

2. Set up equations:

For the bus:

For the car:

3. Solve the equations.

Seemed to work out. Thanks, Bob! I really appreciate it. I can usually solve the problems, but sometimes it's good with a little confirmation that I've done it right, or getting a hint or seeing the proper solution This really, really helps, especially because I don't have many other clever heads to discuss the problems with.

Thanks again!

Thanks for joining in, Bob! That looks awesome.

Note that

Sorry for not mentioning that earlier!

**Hixy**- Replies: 4

Given the function

withFind the real function g(t) and imaginary function h(t) of f(t).

My solution:

Is this correct?

2nd question:

Find

My solution:

or

You guys agree? I'm mostly in trouble with the first part. Note that the vector line is just the sign of the complex conjugate, didn't know how to make it just a line.

Since

is known to be a solution, by deflating the polynomial we get.

By inspecting the 3rd degree polynomial, we see that another root is 2. This can also be found by graphing it and find the intersection with the x-axis.

Thus the problem is already solved. 2 is a real root with multiplicity 2 and a(1+i) and a(1-i) are the complex roots.

However, deflating the 3rd degree polynomial gives

Taking the quadratic of the 2nd degree polynomial we get

.

So the two complex roots are -1+i and -1-i. Therefore a=1.

**Hixy**- Replies: 7

Hello everybody

**What kind of help I'm looking for:** A hint to where it goes wrong in my solution. A step by step solution that I can look at to verify that my work is correct after I've used the hint. Or any other kind of help you can give me.

**What kind of level I'm at:** Go ahead with the hard explanations.

I have the following problem due tomorrow. I know the steps, but doing the actual computation seems tremendously hairy to me and afterwards I can't seem to get a desired result.

The problem:

We are given the complex function

The problem statement gives 2 of the 4 roots as

and

My current thoughts:

The third root is obviously the complex conjugate:

Now my job is to find the last root.

Method: Factorize the entire polynomium with 1 unknown root

and isolate. However, what becomes my problem is that the a's from the complex numbers won't go away, and then I have 2 unknowns.My try at a solution:

It was used that

and

Where do I go from here? Or should I have taken an entirely different road?

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